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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 84 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 84  Maths Textbook Solution.

Answers (1)

Answer:

\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C

Hint:

You must know about integration of \int \sqrt{a^{2}+x^{2}}

Given:

\int \sqrt{a^{2}+x^{2}} d x

Solution:

I=\int 1_{I I} \cdot \sqrt{a_{1}^{2}+x^{2}} d x

=\sqrt{a^{2}+x^{2}} \int 1 d x-\int\left(\frac{d}{d x}\left(\sqrt{a^{2}+x^{2}}\right) \int 1 d x\right) d x

=\sqrt{a^{2}+x^{2}} \cdot x-\int \frac{1 \times 2 x}{2 \sqrt{a^{2}+x^{2}}} \cdot x d x

=\sqrt{a^{2}+x^{2}} \cdot x-\int\left(\frac{x^{2}+a^{2}-a^{2}}{\sqrt{a^{2}+x^{2}}}\right) d x

=x \sqrt{a^{2}+x^{2}}-\int \sqrt{a^{2}+x^{2}} d x+a^{2} \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x

=x \sqrt{a^{2}+x^{2}}-I+a^{2} \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x

\therefore 2 I=x \sqrt{a^{2}+x^{2}}+a^{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|

\Rightarrow I=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C

 

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